design of slow sand filter final

Adavisomanal and 30 Villages Jal NIrmal Project

Detalied Scheme Report Niketan Consultants

Design of Slow sand filterAdavisomanal & 30 Villages

As per CPHEEO manual clause the economical cost design of slow sand filter to be adopted

Total quantity of water per day to be treated 3600 Cum / day

Since slow sand filters work for 24 hrs a day the quantity to be filter per hour

= 3600 = 150 Cum / per hour24

The rate of filtration for slow sand varies between 0.1 M to 0.2 M per hour. Since sedimentation process followed by Raw water storage tank (3 hrs - storage) has been provided, the rate of filtration in this can be adopted 0.150 M / hr.

= 150 = 1000 Sqm0.150

As per CPHEEO manual 6 Nos of filter beds are to be provided.As per economical cost design (page - 241)

= 2 A Wheren + 1 l = Length of wall

b = breadth of wallb = ( n + 1 ) l A = Total surface area of filter bed

2 n n = Nos of filter beds

A = n l b

As surface works out Less than 1200 Sqm.As per CPHEEO manual, number filter beds required is 5 But provide 8 Nos.

Therefore, A = 8 l x b

= 2 x 10008 + 1

= 2000 = 222.222222229

l = 14.90711985Say = 15.00 M

Therefore, A = 8 x l x b

1000 = 8 x 15.00 x bb = 1000

15 x 8b = 8.3333333333

Say b = 8.5 M

Therefore, The area of slow sand filter bed requirement

l 2

l 2

l 2



Therefore,Area provided is 15.00 x 8.5 m 8 No of BedsTherefore, Each bed will have Length = 15 M

breadth = 8.5 M

Surface area of each bed = 127.5 Sqm

Total Surface area of 8 bed = 127.5 x 8 = 1020 > 1000 Sqm

Hence OkCheak for Over loading

Let us assume 1 Filter bed is Washed at a time

Filteration capacity of each filter = 150 = 0.2 Cum /hr = .2 Cum /hr15 X 8.5 x 7

Hence Ok

The schematic lay out of slow sand filter bed be as under

Design of Filter Bed

Let filter bed be as under with over all depth of filter bed = 2.70 mt

0.20mt Free board

1.0mt supernatant water

2.71.0mt Initial loading of filter Sand

0.40 mt Gravel

0.10 mt Under drain

Sand Spcification

Size of sand = 0.2- 0.3 mm

Uniformity co-eff = 5

Graval Spcificat = 5.0 layers (0.4 m)

Size of garval = 2.0 mm 40 mm

Size of garval i = 2.0 5 10 20 40

depth in mm = 80 80 80 80 80



Design of under Drain

Area of each filter = 15.00 x 8.5 = 127.5 Sq Mts

ratio of area of perforation to area of filter = 0.15% to 0.3 %

Provide area of perforation = 0.15 / 100 x 127.5

= 0.19125 Sqm 1912.5 Sqcm

Ratio of total perforation to total C/S area of latral

for perforation of 12 mm = 2 to 1 % cross sectional area

Total Cross sectional area of latrals = 1.5 x 1913

= 2868.8 Sqcmt

Area of central manifold = 1.5 x 2868.75

= 4303.125 Sqcmt

Cross section of center manifold with depth of 100 cmt

therefore width of central manifold = 4303 Cm

Provide central manifold of 60 cmt depth ,75 cmt wide

Design of Lateral:-

Assuming 63 mm dia of PVC cmt latral

Now internal Dia of 63 mm dia PVC Pipe = 58.1 mm

No. of Laterals = Total Area of laterls

C/S Area of one Lateral

= 28.6875 = 110 Nos.

Spacing of Laterals= 15 = 0.273 m

55 Say 250mm C/C

Length of Laterals on each side = Width of filter bed - Width of Manifold

2

= 8.5 - 0.75 = 3.875m

2

Which is Less Than 3.78(60x.063=3.78m) Hence Ok

Design of Orifice:-

Area of Orifice = 0.2 X 15 X 8.5

100

= 0.19125 Sqm

Adopting Spacing Of 18 cm C/C

Total Area of Orifice on each Laterals = 3.875 x 100

= 387.5 Cm

No. of Orifice = 0.1913 x 100 x 100 = 9 Nos.

110 x 2

Diameter of Orifice d= 0.19125 x 4 = 0.49 Cm

3.142

0.785x0.05812



Thus provide 9mm Orifice at 180 mm C/C

Adavisomanal And 30 Villages Jal Nirmal Project

Detailed Scheme Report Niketan Consultants

Design Of Sedimentation tank Total demand = 3600 Cum/Day

=With 20 hrs pumping = 180 Cum/Day

Detention provide = 3 Hrs

Capacity Sedimentation Tank = 3 X 180 = 540 Cum

If two units provided

capacity of each unit =

540= 270 Cum

2

with average water depth in tank as 3.0 mts, Surface area of tank = 270

= 90 Sqm3

Length to Width ratio is 1 : 3 = 3B2 = 90 SqmB = 5.48Say 5.5 M

L = 3 x 5.50 = 16.5MtsSo, Provide 16.5 Mts X 5.5 Mts of each unit Total Area Provided = 90.75 Sqm > 90 Sqm Ok

Total area provided = 2 (16.5 x 5.5 x 3.0) = 544.5cum > 540 Cum Hence, OK

16.5 Mts

Unit 15.5 Mts

Unit 2 5.5 Mts

16.5 Mts

Check for Surface loading = 3600

= 19.840 Cum/Sqm/Day < 36 Cum/Sqm/Day90.75 x 2.0

Hence Ok

Overall Depth of each Unit with 3.0 mts standing water depth (SWD) = 3 + 0.3 0 (FB) + 0.3 (Sluddge) += 3.6 Mts

Inlet And out let Weir

Inlet weir (Influent channel with bottom openings )

keeping velocity of water in channel = 0.3 m/sec

and keeping width of channel as = 0.3 m

For discharge of 180 Cum/hr i.e. = 0.05 Cum/sec

Depth of channel in each unit = 1/2 x

0.05 = 0.28 Mts0.3 x 0.3 Say 30 cmt

With FB of 20 cmt , total depth of weir be = 50 cmt



The cross section of Influent channel = 0.3x 0.3 mCheck for weir loading Length of wier in eah unit = 5.5 MtsFlow per unit per day = 3600 /2 = 180 Cum /hrTherefore weir loading = 1800/ 5.5 = 327.27 cum / mt /day Between 100 -300 cum /mt / day As it ismore 300Cum/m/ day Let us incrseth length of weir to 6.5 m Now The Modified dimension for the Sedimentation Tank is Length of Sedimentation Tank = Area of C/s

=90

=13.846

Breadth 6.5Say 14 M

Check for weir loading Length of wier in eah unit = 6.5mFlow per unit per day = 3600 /2 = 180 Cum/hrTherefore weir loading = 1800/ 6.5 = 276.93 cum / mt /day Between 100 -300 cum /mt / day

Hence Ok

Check for Surface loading = 3600

= 19.780 Cum/Sqm/Day < 36 Cum/Sqm/Day91 x 2.0

Hence OkThus the Final Dimension Are = 2 Sets 14 m x 6.5m

14.0 Mts

6.5 Mts

6.5 Mts

14.0 Mts

Bottom floor slope 1 : 12

Hydraulic Design of baffle wall

Flow in each Tank = 180 Cum/hr = 0.05 Cum/Sec

Velocity of Flow of flow of water for dispersal = 0.15 /Sec

Area of opening required – in baffle wall = 0.15 = 0.333 Sqm

3 0.15

Opening below baffle wall for movement of sludge and water, provided 45

cm considering sludge deposit of 25 cm before flushed one.

Therefore opening available below for flow of water (45-25)_ 20 Cm

The baffle wall is supported over 20 x 20 Cm columns 3 Nos.

Hence length of opening available = 6.25 - (0.2x3)

= 5.65 M

Area of opening available = 5.65 x 0.2 = 1.13 Sqm.

Hence bottom opening adequate.

But to make proper dispersal at higher level

For Sludge removal, let is divide the Box Slab in each tank into 2 halves longitudinally and slope be provided in word to make into to two hopper of 30 x 30 m in the center and sludge removed by by sludge pipe. To economies construction cost, let the tanks be constructed back to back longitudinally to ensure equal flow into sedimentation tank a distribution box with rectangular weir and flush type inlet valve to each tank.

For uniform dispersal of flow of water provide perforated baffle wall in front of inlet pipe at 1.2 M away from the wall.



Let us provide 15 x 15 Cm Square opening in the wall as under.

Adavisomanal and 30 Villages Jal Nirmal Project


Design of Pipe from Raw Water Sump To Distribution BoxNo. of Filter Beds= 8 NosQty of water to be Treated= 150 Cum/HrQty of water to be Treated by pipe= 150 Cum/Hr

= 0.04167 Cum/SecAssuming the Velocity of flow In the pipe as= 0.5 M/sec

Dia of pipe= 0.04167 x 40.5 x 3.142

= 0.3257Provide 250 mm Dia CI Pipe

Now, Velocity in pipe V=Q/AV= 0.04167

0.785 X 0.25 X 0.25= 0.85 M/sec

Hence OkCalculation Of Head Losses in pipe:

Length of pipe= 30 mHazen Willam Constant= 100

Area of Cross Section in pipe A= 3.142 X 250 X 250 = 490944

Wetted perimeter = 3.142 x 250 = 785.5 mmHydraulic Mean Radius r = 62.5 mm = 0.0625 m

Now, Accroding to Hazen Willams Formula 0.63 0.54

0.85 = 0.854 x 100 x 0.0625 x SS = 0.00498 m/m

Frictional Losses= 0.164 mOut let RL at Raw water Sump= 595.2 Minlet RL at Distribution Chamber= 595.036 M

Design Of interconnecting From Distribution tank(Inlet Equidistrbution Retangular Weir) For Slow sand Filters Since Separate pipes Are Tobe Give From Distribution Tank to SSFDischage In each Filter= 0.00521 Cum/SecAssuming the Velocity of flow In the pipe as= 0.5 M/sec

Dia of pipe= 0.00521 x 40.5 x 3.142



0.785 X 0.1 X 0.1= 0.66

Hence OkCalculation Of Head Losses in pipe:For SSF 1,SSF 4 on either side



Wetted perimeter = 3.142 x 100 = 314.2 mmHydraulic Mean Radius r = 25 mm = 0.025 m


0.66 = 0.854 x 100 x 0.025 x SS = 0.00498 m/m

Frictional Losses= 0.16434 mOut let RL at Distribution Box= 595.03 Minlet RL at Distribution Chamber= 594.86566 M

mm2

V= Q/A=0.854XCX r.63xS0.54

mm2

V= Q/A=0.854XCX r.63xS0.54



For SSF 2,SSF 3 on either sideLength of pipe= 15 m

Hazen Willam Constant= 100




M = 0.854 x 100 x 0.025 x SS = 0.00498 m/m

Frictional Losses= 0.08217 mOut let RL at Distribution Box= 595.03 Minlet RL at Distribution Chamber= 594.948 M

Design Of interconnecting From Slow sand Filters Out let to Rure Water Sump One of Slow Sand Filter = 8 Nos.Qty of water to be Treated= 143.75 Cum/HrQty of water to be Treated by each pipe= 17.9688 Cum/Hr

= 0.00499 Cum/Seca. From SSF 1 to Junction SSF2Since pipes Are inter connceted from SSF 1 to SSF2Discharge in pipe = 0.0049913 Cum/SecAssuming the Velocity of flow In the pipe as= 0.5 M/sec

Dia of pipe= 0.00499 x 40.5 x 3.142



0.785 X 0.1 X 0.1

= 0.636 M/secHence Ok

Calculation Of Head Losses in pipe:





0.64 = 0.854 x 100 x 0.025 x SS = 0.00498 m/m

Frictional Losses= 0.08217 mOut let RL at Slow Sand Filter = 593.88 M RL at Outlet of SSF 2 = 593.798 M

b. From SSF 2 to Junction SSF3Since pipes Are inter connceted from SSF 2to SSF3Discharge in pipe = 0.0099826 Cum/SecAssuming the Velocity of flow In the pipe as= 0.5 M/sec

Dia of pipe= 0.00998 x 40.5 x 3.142

= 0.1594

mm2

V= Q/A=0.854XCX r.63xS0.54

mm2

V= Q/A=0.854XCX r.63xS0.54



Provide 150 mm Dia CI Pipe Now, Velocity in pipe V=Q/A

V= 0.009980.785 X 0.15 X 0.15







0.57 = 0.854 x 100 x 0.0375 x SS = 0.00498 m/m

Frictional Losses= 0.08217 m RL at Outlet of SSF 2 = 593.798 M RL at Outlet of SSF 3= 593.716 M

c. From SSF 3 to Junction SSF4Since pipes Are inter connceted from SSF3 to SSF4Discharge in pipe = 0.014974 Cum/SecAssuming the Velocity of flow In the pipe as= 0.5 M/sec

Dia of pipe= 0.01497 x 40.5 x 3.142



0.785 X 0.15 X 0.15







0.85 = 0.854 x 100 x 0.0375 x SS = 0.00498 m/m

Frictional Losses= 0.16434 mOut let RL at Raw water Sump= 593.716 Minlet RL at Distribution Chamber= 593.551 M

mm2

V= Q/A=0.854XCX r.63xS0.54

mm2

V= Q/A=0.854XCX r.63xS0.54



d. From SSF 4 to pure Water SumpSince pipes Are inter connceted from SSF3 to SSF4Discharge in pipe = 0.0199653 Cum/SecAssuming the Velocity of flow In the pipe as= 0.5 M/sec

Dia of pipe= 0.01997 x 40.5 x 3.142



0.785 X 0.2 X 0.2= 0.636 M/sec

Hence OkCalculation Of Head Losses in pipe:





0.64 = 0.854 x 100 x 0.05 x SS = 0.00498 m/m

Frictional Losses= 0.16434 m RL at Outlet of SSF 3 = 593.551 M RL at inlet Level of Sump = 593.387 M

mm2

V= Q/A=0.854XCX r.63xS0.54

design of slow sand filter final

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